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Distribution of-Machine Friction.

Next plot these figures as in Fig. 595, the horizontal
shewing W, and the vertical ordinates the corresponding va
P; o D being the theoretical, and A B the practical profiles, "**"'

are both straight lines.    Draw AC || to o D.    Then at any ore!i
except o A the total P consists of:

1.  Force to overcome load, neglecting friction.

2.  Force to overcome friction of unloaded machine.

3.  Force to overcome friction due to load.

(2) being a constant quantity as shewn between lines AC, o f J

Then, if iv be the equivalent weight of the unloaded
causing friction,

(i)          (2)        (3)

F            BC

From (3) we find u = -~ =----------= '00687.

r     W      5 x 2240             '

Supposing p a constant throughout the machine,


p ----+ -00687 w 4- -00687 W


and as CA = 23*1 = 3362 x -00687     :    w  3362 Ibs.